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Byju's Answer
Standard XII
Mathematics
Trigonometric Ratios of Compound Angles
Q27 27. ...
Question
Q27
27
.
I
f
a
cos
3
θ
+
3
a
sin
2
θ
cos
θ
=
m
a
n
d
a
sin
3
θ
+
3
a
sin
θ
cos
2
θ
=
n
,
p
r
o
v
e
t
h
a
t
m
+
n
2
/
3
+
m
-
n
2
/
3
=
2
a
2
/
3
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Solution
Dear student
Given
:
m
=
acos
3
θ
+
3
asin
2
θ
cosθ
n
=
asin
3
θ
+
3
asinθ
cos
2
θ
Consider
,
LHS
m
+
n
2
3
+
m
-
n
2
3
=
acos
3
θ
+
3
asin
2
θ
cosθ
+
asin
3
θ
+
3
asinθ
cos
2
θ
2
3
+
acos
3
θ
+
3
asin
2
θ
cosθ
-
asin
3
θ
-
3
asinθ
cos
2
θ
2
3
=
a
2
3
cos
3
θ
+
sin
3
θ
+
3
cosθ
sinθ
cosθ
+
sinθ
2
3
+
a
2
3
cos
3
θ
-
sin
3
θ
-
3
cosθ
sinθ
cosθ
-
sinθ
2
3
=
a
2
3
cosθ
+
sinθ
3
2
3
+
a
2
3
cosθ
-
sinθ
3
2
3
=
a
2
3
cos
2
θ
+
sin
2
θ
+
2
sinθcosθ
+
cos
2
θ
+
sin
2
θ
-
2
sinθcosθ
=
a
2
3
×
2
×
cos
2
θ
+
sin
2
θ
=
2
a
2
3
=
RHS
Identity
used
:
1
)
cos
2
x
+
sin
2
x
=
1
Regards
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0
Similar questions
Q.
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If
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sin
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θ
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If θ =
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Q.
If a cos
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